# Question on graph-measurability of a random set

Example 1.3.34 on page 74 of Ilya Molchanov's Theory of Random Sets makes the claim that if $$(\xi(\omega,t);\omega\in\varOmega,t\geq0)$$ is a (jointly measurable) stochastic process, then its path $$\varXi = \{\xi(\cdot,t)\colon t\geq0\}$$ is graph measurable.

By graph measurable, the author means that the set $$\{(\omega,r)\in\varOmega\times\mathbb{R}\colon r\in\varXi(\omega)\}$$ is in the product $$\sigma$$-algebra $$\mathcal{A}\otimes\mathcal{B}(\mathbb{R})$$, where $$\mathcal{A}$$ is a $$\sigma$$-algebra on the sample space $$\varOmega$$ and $$\mathcal{B}(\mathbb{R})$$ is the Borel $$\sigma$$-algebra on $$\mathbb{R}$$.

This is how I tried proceeding:

$$\{(\omega,r)\in\varOmega\times\mathbb{R}\colon r\in\varXi(\omega)\} = \{(\omega,r)\in\varOmega\times\mathbb{R}\colon \exists t\geq0~\text{such that}~\xi(\omega,t) = r\} = \bigcup_{t\geq0}\{(\omega,r)\in\varOmega\times\mathbb{R}\colon \xi(\omega,t) = r\}.$$

Now, for each fixed $$t$$, the set $$\{(\omega,r)\in\varOmega\times\mathbb{R}\colon \xi(\omega,t) = r\}$$ is easily shown to be in $$\mathcal{A}\otimes\mathcal{B}(\mathbb{R})$$. But, without any further assumptions, I don't really see why that uncountable union over the $$t$$-s would also be in the product $$\sigma$$-algebra.

And, in addition, if this claim is not true in general, what assumptions (path continuity or, may be, separability??) can be imposed on the process $$\xi$$ to make it happen?

I can tell you it is not true in general.

Counterexample

Let $$\mathcal{A}=\{\emptyset,\Omega\}$$ and define the measurable function $$f:[0,1]\to[0,1]$$ be as in this answer, whose image $$E=\{f(t):t\in[0,1]\}$$ is not measurable. Define, for all $$\omega\in\Omega$$, $$\xi(\omega,t)=f(\max\{t,1\})$$. The measurability of $$f$$ implies that of $$\xi$$. However, $$\Xi(\omega)=\{f(t):t\in[0,1]\}=E$$ and $$\{(\omega,r)\in\Omega\times\mathbb{R} :r\in\Xi(\omega)\}=\Omega\times E$$, which is not measurable.

Sufficient conditions

Let $$\mathbb{P}$$ be a probability measure. If $$\xi(\omega,\cdot)$$ is separable on $$\Omega\setminus\Omega_0$$ where $$\Omega_0$$ is $$\mathbb{P}$$-null, and if $$\mathcal{A}$$ contains all subsets of $$\Omega_0$$, then a modification of the closure $$\overline{\Xi}$$ is graph-measurable.

Indeed, let $$\xi'$$ be the indistinguishable from $$\xi$$, defined so that they agree on $$\Omega$$ and $$\xi'$$ is constant and equal to $$0$$ on $$\Omega_0$$ (note that $$\xi'$$ remains jointly measurable, by assumption). Let $$T\subset[0,\infty)$$ be the countable set that separates $$\xi$$. Define, for $$n\geq 1$$ and $$k\in\mathbb{Z}$$, the interval $$I_{k,n}=\big[\frac{k-1}{2^n},\frac{k+1}{2^n}\big]$$. Note that

$$A:=\{(\omega,r)\in\Omega\times\mathbb{R}: r\in\overline{\Xi'(\omega)}\}= \bigcap_{n\in\mathbb{N}}\bigcup_{k\in\mathbb{Z}}\bigcup_{t\in T} \big\{\omega\in\Omega:\big|\xi'(\omega,t)-k2^{-n}\big|\leq 2^{-n}\big\}\times I_{k,n}.$$

(Indeed, $$r$$ is arbitrarily close to some value of $$\xi'(\omega,\cdot)$$ if and only if it is arbitrarily close to (at most $$2^{-n}$$ for any $$n$$) some diadic rational (here $$k2^{-n}$$) which is itself just as close to some value of $$\{\xi'(\omega,t):t\in T\}$$.) The measurability of $$A$$ is clear since (I) these are countable unions and intersection and (II) $$\big\{\omega\in\Omega: \big|\xi'(\omega,t)-k2^{-n}\big|\leq 2^{-n}\big\}$$ is measurable (which follows easily from the joint measurability of $$\xi'$$) and hence so are the rectangles.

I hope this helps.